3.949 \(\int \frac{(d+e x)^m (f+g x)^2}{\sqrt{a+b x+c x^2}} \, dx\)

Optimal. Leaf size=502 \[ \frac{(d+e x)^{m+1} \sqrt{1-\frac{2 c (d+e x)}{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}} \sqrt{1-\frac{2 c (d+e x)}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}} F_1\left (m+1;\frac{1}{2},\frac{1}{2};m+2;\frac{2 c (d+e x)}{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e},\frac{2 c (d+e x)}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right ) \left (e g^2 (m+1) (b d-a e)+c \left (d^2 g^2-2 d e f g (m+2)+e^2 f^2 (m+2)\right )\right )}{c e^3 (m+1) (m+2) \sqrt{a+b x+c x^2}}-\frac{g (d+e x)^{m+2} \sqrt{1-\frac{2 c (d+e x)}{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}} \sqrt{1-\frac{2 c (d+e x)}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}} (b e g (2 m+3)+c (2 d g-4 e f (m+2))) F_1\left (m+2;\frac{1}{2},\frac{1}{2};m+3;\frac{2 c (d+e x)}{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e},\frac{2 c (d+e x)}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{2 c e^3 (m+2)^2 \sqrt{a+b x+c x^2}}+\frac{g^2 \sqrt{a+b x+c x^2} (d+e x)^{m+1}}{c e (m+2)} \]

[Out]

(g^2*(d + e*x)^(1 + m)*Sqrt[a + b*x + c*x^2])/(c*e*(2 + m)) + ((e*(b*d - a*e)*g^2*(1 + m) + c*(d^2*g^2 + e^2*f
^2*(2 + m) - 2*d*e*f*g*(2 + m)))*(d + e*x)^(1 + m)*Sqrt[1 - (2*c*(d + e*x))/(2*c*d - (b - Sqrt[b^2 - 4*a*c])*e
)]*Sqrt[1 - (2*c*(d + e*x))/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)]*AppellF1[1 + m, 1/2, 1/2, 2 + m, (2*c*(d + e*
x))/(2*c*d - (b - Sqrt[b^2 - 4*a*c])*e), (2*c*(d + e*x))/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)])/(c*e^3*(1 + m)*
(2 + m)*Sqrt[a + b*x + c*x^2]) - (g*(b*e*g*(3 + 2*m) + c*(2*d*g - 4*e*f*(2 + m)))*(d + e*x)^(2 + m)*Sqrt[1 - (
2*c*(d + e*x))/(2*c*d - (b - Sqrt[b^2 - 4*a*c])*e)]*Sqrt[1 - (2*c*(d + e*x))/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*
e)]*AppellF1[2 + m, 1/2, 1/2, 3 + m, (2*c*(d + e*x))/(2*c*d - (b - Sqrt[b^2 - 4*a*c])*e), (2*c*(d + e*x))/(2*c
*d - (b + Sqrt[b^2 - 4*a*c])*e)])/(2*c*e^3*(2 + m)^2*Sqrt[a + b*x + c*x^2])

________________________________________________________________________________________

Rubi [A]  time = 0.681, antiderivative size = 500, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.138, Rules used = {1653, 843, 759, 133} \[ \frac{(d+e x)^{m+1} \sqrt{1-\frac{2 c (d+e x)}{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}} \sqrt{1-\frac{2 c (d+e x)}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}} F_1\left (m+1;\frac{1}{2},\frac{1}{2};m+2;\frac{2 c (d+e x)}{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e},\frac{2 c (d+e x)}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right ) \left (g^2 (b d-a e)+\frac{c \left (d^2 g^2-2 d e f g (m+2)+e^2 f^2 (m+2)\right )}{e (m+1)}\right )}{c e^2 (m+2) \sqrt{a+b x+c x^2}}-\frac{g (d+e x)^{m+2} \sqrt{1-\frac{2 c (d+e x)}{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}} \sqrt{1-\frac{2 c (d+e x)}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}} (b e g (2 m+3)+2 c d g-4 c e f (m+2)) F_1\left (m+2;\frac{1}{2},\frac{1}{2};m+3;\frac{2 c (d+e x)}{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e},\frac{2 c (d+e x)}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{2 c e^3 (m+2)^2 \sqrt{a+b x+c x^2}}+\frac{g^2 \sqrt{a+b x+c x^2} (d+e x)^{m+1}}{c e (m+2)} \]

Antiderivative was successfully verified.

[In]

Int[((d + e*x)^m*(f + g*x)^2)/Sqrt[a + b*x + c*x^2],x]

[Out]

(g^2*(d + e*x)^(1 + m)*Sqrt[a + b*x + c*x^2])/(c*e*(2 + m)) + (((b*d - a*e)*g^2 + (c*(d^2*g^2 + e^2*f^2*(2 + m
) - 2*d*e*f*g*(2 + m)))/(e*(1 + m)))*(d + e*x)^(1 + m)*Sqrt[1 - (2*c*(d + e*x))/(2*c*d - (b - Sqrt[b^2 - 4*a*c
])*e)]*Sqrt[1 - (2*c*(d + e*x))/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)]*AppellF1[1 + m, 1/2, 1/2, 2 + m, (2*c*(d
+ e*x))/(2*c*d - (b - Sqrt[b^2 - 4*a*c])*e), (2*c*(d + e*x))/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)])/(c*e^2*(2 +
m)*Sqrt[a + b*x + c*x^2]) - (g*(2*c*d*g - 4*c*e*f*(2 + m) + b*e*g*(3 + 2*m))*(d + e*x)^(2 + m)*Sqrt[1 - (2*c*
(d + e*x))/(2*c*d - (b - Sqrt[b^2 - 4*a*c])*e)]*Sqrt[1 - (2*c*(d + e*x))/(2*c*d - (b + Sqrt[b^2 - 4*a*c])*e)]*
AppellF1[2 + m, 1/2, 1/2, 3 + m, (2*c*(d + e*x))/(2*c*d - (b - Sqrt[b^2 - 4*a*c])*e), (2*c*(d + e*x))/(2*c*d -
(b + Sqrt[b^2 - 4*a*c])*e)])/(2*c*e^3*(2 + m)^2*Sqrt[a + b*x + c*x^2])

Rule 1653

Int[(Pq_)*((d_.) + (e_.)*(x_))^(m_.)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Expon[Pq
, x], f = Coeff[Pq, x, Expon[Pq, x]]}, Simp[(f*(d + e*x)^(m + q - 1)*(a + b*x + c*x^2)^(p + 1))/(c*e^(q - 1)*(
m + q + 2*p + 1)), x] + Dist[1/(c*e^q*(m + q + 2*p + 1)), Int[(d + e*x)^m*(a + b*x + c*x^2)^p*ExpandToSum[c*e^
q*(m + q + 2*p + 1)*Pq - c*f*(m + q + 2*p + 1)*(d + e*x)^q - f*(d + e*x)^(q - 2)*(b*d*e*(p + 1) + a*e^2*(m + q
- 1) - c*d^2*(m + q + 2*p + 1) - e*(2*c*d - b*e)*(m + q + p)*x), x], x], x] /; GtQ[q, 1] && NeQ[m + q + 2*p +
1, 0]] /; FreeQ[{a, b, c, d, e, m, p}, x] && PolyQ[Pq, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2
, 0] &&  !(IGtQ[m, 0] && RationalQ[a, b, c, d, e] && (IntegerQ[p] || ILtQ[p + 1/2, 0]))

Rule 843

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dis
t[g/e, Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + b*x + c*x^
2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&&  !IGtQ[m, 0]

Rule 759

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c,
2]}, Dist[(a + b*x + c*x^2)^p/(e*(1 - (d + e*x)/(d - (e*(b - q))/(2*c)))^p*(1 - (d + e*x)/(d - (e*(b + q))/(2
*c)))^p), Subst[Int[x^m*Simp[1 - x/(d - (e*(b - q))/(2*c)), x]^p*Simp[1 - x/(d - (e*(b + q))/(2*c)), x]^p, x],
x, d + e*x], x]] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] &
& NeQ[2*c*d - b*e, 0] &&  !IntegerQ[p]

Rule 133

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[(c^n*e^p*(b*x)^(m +
1)*AppellF1[m + 1, -n, -p, m + 2, -((d*x)/c), -((f*x)/e)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, e, f, m, n, p},
x] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])

Rubi steps

\begin{align*} \int \frac{(d+e x)^m (f+g x)^2}{\sqrt{a+b x+c x^2}} \, dx &=\frac{g^2 (d+e x)^{1+m} \sqrt{a+b x+c x^2}}{c e (2+m)}+\frac{\int \frac{(d+e x)^m \left (\frac{1}{2} e \left (2 c e f^2 (2+m)-g^2 (b d+2 a e (1+m))\right )-\frac{1}{2} e g (2 c d g-4 c e f (2+m)+b e g (3+2 m)) x\right )}{\sqrt{a+b x+c x^2}} \, dx}{c e^2 (2+m)}\\ &=\frac{g^2 (d+e x)^{1+m} \sqrt{a+b x+c x^2}}{c e (2+m)}-\frac{(g (2 c d g-4 c e f (2+m)+b e g (3+2 m))) \int \frac{(d+e x)^{1+m}}{\sqrt{a+b x+c x^2}} \, dx}{2 c e^2 (2+m)}+\frac{\left (e (b d-a e) g^2 (1+m)+c \left (d^2 g^2+e^2 f^2 (2+m)-2 d e f g (2+m)\right )\right ) \int \frac{(d+e x)^m}{\sqrt{a+b x+c x^2}} \, dx}{c e^2 (2+m)}\\ &=\frac{g^2 (d+e x)^{1+m} \sqrt{a+b x+c x^2}}{c e (2+m)}-\frac{\left (g (2 c d g-4 c e f (2+m)+b e g (3+2 m)) \sqrt{1-\frac{d+e x}{d-\frac{\left (b-\sqrt{b^2-4 a c}\right ) e}{2 c}}} \sqrt{1-\frac{d+e x}{d-\frac{\left (b+\sqrt{b^2-4 a c}\right ) e}{2 c}}}\right ) \operatorname{Subst}\left (\int \frac{x^{1+m}}{\sqrt{1-\frac{2 c x}{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e}} \sqrt{1-\frac{2 c x}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}}} \, dx,x,d+e x\right )}{2 c e^3 (2+m) \sqrt{a+b x+c x^2}}+\frac{\left (\left (e (b d-a e) g^2 (1+m)+c \left (d^2 g^2+e^2 f^2 (2+m)-2 d e f g (2+m)\right )\right ) \sqrt{1-\frac{d+e x}{d-\frac{\left (b-\sqrt{b^2-4 a c}\right ) e}{2 c}}} \sqrt{1-\frac{d+e x}{d-\frac{\left (b+\sqrt{b^2-4 a c}\right ) e}{2 c}}}\right ) \operatorname{Subst}\left (\int \frac{x^m}{\sqrt{1-\frac{2 c x}{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e}} \sqrt{1-\frac{2 c x}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}}} \, dx,x,d+e x\right )}{c e^3 (2+m) \sqrt{a+b x+c x^2}}\\ &=\frac{g^2 (d+e x)^{1+m} \sqrt{a+b x+c x^2}}{c e (2+m)}+\frac{\left (e (b d-a e) g^2 (1+m)+c \left (d^2 g^2+e^2 f^2 (2+m)-2 d e f g (2+m)\right )\right ) (d+e x)^{1+m} \sqrt{1-\frac{2 c (d+e x)}{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e}} \sqrt{1-\frac{2 c (d+e x)}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}} F_1\left (1+m;\frac{1}{2},\frac{1}{2};2+m;\frac{2 c (d+e x)}{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e},\frac{2 c (d+e x)}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{c e^3 (1+m) (2+m) \sqrt{a+b x+c x^2}}-\frac{g (2 c d g-4 c e f (2+m)+b e g (3+2 m)) (d+e x)^{2+m} \sqrt{1-\frac{2 c (d+e x)}{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e}} \sqrt{1-\frac{2 c (d+e x)}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}} F_1\left (2+m;\frac{1}{2},\frac{1}{2};3+m;\frac{2 c (d+e x)}{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e},\frac{2 c (d+e x)}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{2 c e^3 (2+m)^2 \sqrt{a+b x+c x^2}}\\ \end{align*}

Mathematica [F]  time = 1.27056, size = 0, normalized size = 0. \[ \int \frac{(d+e x)^m (f+g x)^2}{\sqrt{a+b x+c x^2}} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[((d + e*x)^m*(f + g*x)^2)/Sqrt[a + b*x + c*x^2],x]

[Out]

Integrate[((d + e*x)^m*(f + g*x)^2)/Sqrt[a + b*x + c*x^2], x]

________________________________________________________________________________________

Maple [F]  time = 1.701, size = 0, normalized size = 0. \begin{align*} \int{ \left ( ex+d \right ) ^{m} \left ( gx+f \right ) ^{2}{\frac{1}{\sqrt{c{x}^{2}+bx+a}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^m*(g*x+f)^2/(c*x^2+b*x+a)^(1/2),x)

[Out]

int((e*x+d)^m*(g*x+f)^2/(c*x^2+b*x+a)^(1/2),x)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (g x + f\right )}^{2}{\left (e x + d\right )}^{m}}{\sqrt{c x^{2} + b x + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^m*(g*x+f)^2/(c*x^2+b*x+a)^(1/2),x, algorithm="maxima")

[Out]

integrate((g*x + f)^2*(e*x + d)^m/sqrt(c*x^2 + b*x + a), x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (g^{2} x^{2} + 2 \, f g x + f^{2}\right )}{\left (e x + d\right )}^{m}}{\sqrt{c x^{2} + b x + a}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^m*(g*x+f)^2/(c*x^2+b*x+a)^(1/2),x, algorithm="fricas")

[Out]

integral((g^2*x^2 + 2*f*g*x + f^2)*(e*x + d)^m/sqrt(c*x^2 + b*x + a), x)

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**m*(g*x+f)**2/(c*x**2+b*x+a)**(1/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (g x + f\right )}^{2}{\left (e x + d\right )}^{m}}{\sqrt{c x^{2} + b x + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^m*(g*x+f)^2/(c*x^2+b*x+a)^(1/2),x, algorithm="giac")

[Out]

integrate((g*x + f)^2*(e*x + d)^m/sqrt(c*x^2 + b*x + a), x)